Balancing of reciprocating masses
Upcoming SlideShare
Loading in...5
×
 

Balancing of reciprocating masses

on

  • 4,102 views

Unit 5- balancing of reciprocating masses, Dynamics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com ...

Unit 5- balancing of reciprocating masses, Dynamics of machines of VTU Syllabus prepared by Hareesha N Gowda, Asst. Prof, Dayananda Sagar College of Engg, Blore. Please write to hareeshang@gmail.com for suggestions and criticisms.

Statistics

Views

Total Views
4,102
Views on SlideShare
4,055
Embed Views
47

Actions

Likes
7
Downloads
244
Comments
1

4 Embeds 47

http://hareeshang.wikifoundry.com 30
http://www.hareeshang.wikifoundry.com 14
http://allsparklearning.com 2
http://hareeshang.wordpress.com 1

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Balancing of reciprocating masses Balancing of reciprocating masses Presentation Transcript

  • Unit-4 Balancing of Reciprocating Masses Hareesha N Gowda Lecturer Dept of Aeronautical Engg Dayananda Sagar College of Engineering 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 1
  • Unit 4: Balancing of Reciprocating masses • Inertia effect of crank and connecting rod • single cylinder engine • balancing in multi cylinder-inline engine – primary & Secondary forces • V-type engine • Radial engine – Direct and reverse crank method 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 2
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 3
  • Balancing of reciprocating masses- An Introduction • Wkt, various forces acting on the reciprocating parts of an engine. • The resultant of all the forces acting on the body of the engine due to inertia forces only is known as unbalanced force or shaking force. • Thus if the resultant of all the forces due to inertia effects is zero, then there will be no unbalanced force, but even then an unbalanced couple or shaking couple will be present. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 4
  • Balancing of reciprocating masses- An Introduction • Consider a horizontal reciprocating engine mechanism as shown in Fig. FR = Force required to accelerate the reciprocating parts. FI = Inertia force due to reciprocating parts, FN = Force on the sides of the cylinder walls or normal force acting on the cross-head guides, and FB = Force acting on the crankshaft bearing or main bearing. • Since FR and FI, are equal in magnitude but opposite in direction, therefore they balance each other. • The horizontal component of FB (i.e. FBH) acting along the line of reciprocation is also equal and opposite to FI . • This force FBH = FU is an unbalanced force or shaking force and required to be properly balanced. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 5
  • Balancing of reciprocating masses- An Introduction • • • The force on the sides of the cylinder walls (F N) and the vertical component of FB (i.e. FBV) are equal and opposite and thus form a shaking couple of magnitude F N x X or FBV x X. From above we see that the effect of the reciprocating parts is to produce a shaking force and a shaking couple. Since the shaking force and a shaking couple vary in magnitude and direction during the engine cycle, therefore they cause very objectionable vibrations. Thus the purpose of balancing the reciprocating masses is to eliminate the shaking force and a shaking couple. In most of the mechanisms, we can reduce the shaking force and a shaking couple by adding appropriate balancing mass, but it is usually not practical to eliminate them completely. In other words, the reciprocating masses are only partially balanced. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 6
  • Analytical Method for Velocity and Acceleration of the Piston 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 7
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 8
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 9
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 10
  • Primary and Secondary Unbalanced Forces of Reciprocating Masses • Consider a reciprocating engine mechanism as shown in Fig. Let, m = Mass of the reciprocating parts, L = Length of the connecting rod PC, r = Radius of the crank 0C, θ = Angle of inclination of the crank with the line of stroke PO, ω = Angular speed of the crank, n = Ratio of length of the connecting rod to the crank radius = L/r. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 11
  • We have already discussed that, the acceleration of the reciprocating parts is approximately given by the expression. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 12
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 13
  • Partial Balancing of Unbalanced Primary Force In a Reciprocating Engine The primary unbalanced force (m-ffl rcos6) may be considered as the component of the centrifugal force produced by a rotating mass m placed at the crank radius r, as shown in Fig. The primary force acts from O to P along the line of stroke. Hence, balancing of primary force is considered as equivalent to the balancing of mass m rotating at the crank radius r. •This is balanced by having a mass B at a radius b, placed diametrically opposite to the crank pin C. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 14
  • •A little consideration will show, that the primary force is completely balanced if B.b = m.r, but the centrifugal force produced due to the revolving mass B has also a vertical component (perpendicular to the line of stroke) of magnitude • This force remains unbalanced. The maximum value of this force is equal to Bω2b when θ is 90° and 270°, which is same as the maximum value of the primary force mω2r 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 15
  • • From the above discussion, we see that in the first case, the primary unbalanced force acts along the line of stroke whereas in the second case, the unbalanced force acts along the perpendicular to the line of stroke. • The maximum value of the force remains same in both the cases. • It is thus obvious, that the effect of the above method of balancing is to change the direction of the maximum unbalanced force from the line of stroke to the perpendicular of line of stroke. • As a compromise, let a fraction ‘c’ of the reciprocating masses is balanced, such that: cmr = B.b • Therefore, Unbalanced force along the line of stroke 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 16
  • and unbalanced force along the perpendicular to the line of stroke 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 17
  • A single cylinder reciprocating engine has speed 240 r.p.m., stroke 300 mm, mass of reciprocating parts 50 kg, mass of revolving parts at 150 mm radius is 37 kg. If two-third of the reciprocating parts and all the revolving parts are to be balanced, find: 1. The balance mass required at a radius of 400 mm, and 2. The residual unbalanced force when the crank has rotated 60° from top dead centre. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 18
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 19
  • Balancing of Primary Forces of Multi-cylinder In-line Engines • The multi-cylinder engines with the cylinder centre lines in the same plane and on the same side of the centre line of the crankshaft, are known as In-line engines. • The following two conditions must be satisfied in order to give the primary balance of the reciprocating parts of a multi-cylinder engine : 1) The algebraic sum of the primary forces must be equal to zero. In other words, the primary force polygon must close ; and 2) The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero. In other words, the primary couple polygon must close. We have already discussed, that the primary unbalanced force due to the reciprocating masses is equal to the component, parallel to the line of stroke, of the centrifugal force produced by the equal mass placed at the crankpin and revolving with it. Therefore, in order to give the primary balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the reciprocating masses to be transferred to their respective crankpins and to treat the problem as one of revolving masses. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 20
  • Notes : 1) For a two cylinder engine with cranks at 180°, condition (1) may be satisfied, but this will result in an unbalanced couple. Thus the above method of primary balancing cannot be applied in this case. 2) For a three cylinder engine with cranks at 120° and if the reciprocating masses per cylinder are same, then condition (1) will be satisfied because the forces may be represented by the sides of an equilateral triangle. However, by taking a reference plane through one of the cylinder centre lines, two couples with non-parallel axes will remain and these cannot vanish vectorially. Hence the above method of balancing fails in this case also. 3) For a four cylinder engine, similar reasoning will show that complete primary balance is possible and it follows that For a multi-cylinder engine, the primary forces may be completely balanced by suitably arranging the crank angles, provided that the number of cranks are not less than four'. • The closing side of the primary force polygon gives the maximum unbalanced primary force and the closing side of the primary couple polygon gives the maximum un-balanced primary couple. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 21
  • Balancing of Secondary Forces of Multi-cylinder In-line Engines • When the connecting rod is not too long (i.e. when the obliquity of the connecting rod is considered), then the secondary disturbing force due to the reciprocating mass arises. • We have discussed that the secondary force, • As in case of primary forces, the secondary forces may be considered to be equivalent to the component, parallel to the line of stroke, of the centrifugal force produced by an equal mass placed at the imaginary crank of length r/4n and revolving at twice the speed of the actual crank (i.e. 2 ω) as shown in Fig. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 22
  • • Thus, in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to the crankpin is inclined to the line of stroke at twice the angle of the actual crank. • The values of the secondary forces and couples may be obtained by considering the revolving mass. This is done in the similar way as discussed for primary forces. • The following two conditions must be satisfied in order to give a complete secondary balance of an engine : 1. 2. The algebraic sum of the secondary forces must be equal to zero. In other words, the secondary force polygon must close, and The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero. In other words, the secondary couple polygon must close. Note : The closing side of the secondary force polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum unbalanced secondary couple. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 23
  • • In order to give the primary balance of the reciprocating parts of a multicylinder engine, the problem may be treated as that of revolving masses with the reciprocating masses transferred to their respective crank pins. • The position of planes is shown in Fig. a 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 24
  • Assuming the plane of third cylinder as the reference plane, the data may be tabulated as given in Table First of all, the angular position of cranks 2 and 4 are obtained by drawing the couple polygon from the data given in Table (column 6). Assume the position of crank 1 in the horizontal direction as shown in Fig (b), The couple polygon, as shown in Fig. (c), is drawn as discussed below: 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 25
  • •In order to find the mass of the third cylinder (m 3) and its angular position, draw the force polygon, to some suitable scale, as shown in Fig. d, from the data given in Table (column 4). 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 26
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 27
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 28
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 29
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 30
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 31
  • • • • • The secondary crank positions obtained by rotating the primary cranks at twice the angle, is shown in Fig (e). Now draw the secondary force polygon, as shown in Fig. ( f ), to some suitable scale, from the data given in Table (column 4). The closing side of the polygon shown dotted in Fig. ( f ) represents the maximum secondary unbalanced force. By measurement, we find that the maximum secondary unbalanced force is proportional to 582 kg-m. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 32
  • • The position of the cylinders and the cranks are shown in Fig. (a), (b) and (c). 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 33
  • • With the reference plane midway between the cylinders 3 and 4, the data may be tabulated as given in the following table : 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 34
  • • From Fig. (d) and (f), we see that the primary force polygons and couple polygons are closed figures, therefore there are no out-of-balance primary forces. • Thus the engine is balanced for primary forces and couples. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 35
  • • From Fig. (e) and (g), we see that the secondary force polygons and couple polygons are closed figures, therefore there are no out-of-balance secondary forces and couples. • Thus the engine is balanced for secondary forces and couples. Hareesha N G, Dept of Aero Engg, DSCE, Blore 11/27/13 36
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 37
  • Balancing of Radial Engines (Direct and Reverse Cranks Method ) • The method of direct and reverse cranks is used in balancing of radial or V-engines, in which the connecting rods are Connected to a common crank. Since the plane of rotation of the various cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or secondary couple. • Consider a reciprocating engine mechanism as shown in Fig. • Let the crank OC (known as the direct crank) rotates uniformly at ω radians per second in a clockwise direction. Let at any instant the crank makes an angle θ with the line of stroke OP. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 38
  • • The indirect or reverse crank OC' is the image of the direct crank OC, when seen through the mirror placed at the line of stroke. • A little consideration will show that when the direct crank revolves in a clockwise direction, the reverse crank will revolve in the anticlockwise direction. • We shall now discuss the primary and secondary forces due to the mass (m) of the reciprocating parts at P. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 39
  • Considering the primary forces • We have already discussed that primary force is • This force is equal to the component of the centrifugal force along the line of stroke, produced by a mass (m) placed at the crank pin C. • Now let us suppose that the mass (m) of the reciprocating parts is divided into two parts, each equal to m/2. • It is assumed that m/2 is fixed at the direct crank (termed as primary direct crank) pin C and m/2 at the reverse crank (termed as primary reverse crank) pin C', as shown in Fig. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 40
  • • We know that the centrifugal force acting on the primary direct and reverse crank • Hence, for primary effects the mass m of the reciprocating parts at P may be replaced by two masses at C and C' each of magnitude m/2. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 41
  • Note : • The component of the centrifugal forces of the direct and reverse cranks, in a direction perpendicular to the line of stroke, are each equal to , but opposite in direction. • Hence these components are balanced. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 42
  • • In the similar way as discussed above, it will be seen that for the secondary effects, the mass (m) of the reciprocating parts may be replaced by two masses (each m/2) placed at D and D‘ such that OD = OD’ = r/4n. • The crank OD is the secondary direct crank and rotates at 2ω rad/s in the clockwise direction, while the crank OD' is the secondary reverse crank and rotates at 2ω rad/s in the anti CW direction as shown in Fig. . 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 43
  • • The position of three cylinders is shown in Fig. Let the common crank be along the inner dead centre of cylinder 1. Since common crank rotates clockwise, therefore θ is positive when measured clockwise. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 44
  • Maximum primary force acting on the frame of the compressor • The primary direct and reverse crank positions as shown in Fig (a) and (b), are obtained as discussed below : 1) Since θ=0° for cylinder 1, therefore both the primary direct and reverse cranks will coincide with the common crank. 2) Since θ = ±120° for cylinder 2, therefore the primary direct crank is 120° clockwise and the primary reverse crank is 120° anti-clockwise from the line of stroke of cylinder 2. 3) Since θ= ± 240° for cylinder 3, therefore the primary direct crank is 240° clockwise and the primary reverse crank is 240° anti-clockwise from the line of stroke of cylinder 3. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 45
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 46
  • Maximum secondary force acting on the frame of the compressor • The secondary direct and reverse crank positions as shown in Fig. (a) and (b) are obtained as discussed below : 1. Since θ =0° and 2θ =0° for cylinder 1, therefore both the secondary direct and reverse cranks will coincide with the common crank. 2. Since θ = ±120° and 2θ = ± 240° for cylinder 2, therefore the secondary direct crank is 240° clockwise and the secondary reverse crank is 240° anticlockwise from the line of stroke of cylinder 2. 3. Since θ = ± 240° and 2θ = ± 480°, therefore the secondary direct crank is 480° or 120° clockwise and the secondary reverse crank is 480° or 120° anti-clockwise from the line of stroke of cylinder 3. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 47
  • • From Fig.(a), we see that the secondary direct cranks form a balanced system. • Therefore there is no unbalanced secondary force due to the direct cranks. From Fig. (b) we see that the resultant secondary force is equivalent to the centrifugal force of a mass 3 m/2 attached at a crank radius of r/4n and rotating at a speed of 2ω rad/s in the opposite direction to the crank.2 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 48
  • Balancing of V-engines • • Consider a symmetrical two cylinder V-engine as shown in Fig. The common crank OC is driven by two connecting rods PC and QC. The lines of stroke OP and OQ are inclined to the vertical OY, at an angle α. 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 49
  • • The balancing of V-engines is only considered for primary and secondary forces as discussed below : 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 50
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 51
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 52
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 53
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 54
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 55
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 56
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 57
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 58
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 59
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 60
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 61
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 62
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 63
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 64
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 65
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 66
  • 11/27/13 Hareesha N G, Dept of Aero Engg, DSCE, Blore 67